#math/linear-algebra The determinant of the transpose is the determinant of the original. If echelon form of $A^{\top}$ does not have pivots in every column and row, then $A$ is not invertible and $\det A=0.$^[Recall determinant is only defined for square matrices so $A$ is assumed to be square.] When $A$ is invertible, however, echelon form gives a triangular matrix. Replacement operations preserve the determinant. If elimination also uses swaps or scalings, their effects must be accounted for when recovering the original determinant from the triangular matrix's diagonal product.^[Replacement preserves. Swapping negatives. Scaling multiplies.] ## Proposition The $\det A=0$ if and only if $A$ is not invertible. Equivalently $\det A\neq{}0\iff A\text{ is invertible.}\tag{1.1}$ ## Proposition For a square matrix $A$, $ \det A=\det A^{\top}.\tag{1.2} $ ## Proposition For $n\times n$ matrices, $A$ and $B$ $ \det(AB)=(\det A)(\det B), \tag{1.3} $ i.e. the determinant of a product equals the product of of determinants. ## Lemma To prove these, we need something to help us. For a square matrix $A$ and an `elementary matrix`^[An elementary matrix is the matrix of a . [[2B2 - Row Operations as Matrix Multiplications|row reduction operation]].] $E$ of the same size $ \det(AE)=(\det A)(\det E).\tag{1.4} $ The proof is simple. Column operations corresponding to the elementary matrix can be obtained from the identity matrix $I$ by this column operation. So, the determinant is $\det I=1$ times the effect of the column operation. ## Corollary For any matrix $A$ and any sequence of elementary matrices $E_{1},E_{2},\ldots,E_{n}$ with all matrices of shape $n\times n$, then $ \det(AE_1E_2\cdots E_N)=(\det A)(\det E_1)(\det E_2)\ldots(\det E_N)\tag{1.5} $ ## Lemma Any invertible matrix $A$ is a product of elementary matrices. Proof: We know that any invertible matrix is row equivalent to the identity matrix^[Any invertible matrix can be row-operation modified into the identity matrix.] , which is its reduced echelon form. So, $ I=E_{N}E_{N-1}\ldots E_{2}E_{1}A,$ and therefore any invertible matrix can be represented as a product of elementary matrices^[The inverse of an elementary matrix is itself an elementary matrix.] $ A=E_1^{-1}E_2^{-1}\cdots E_{N-1}^{-1}E_N^{-1}I=E_1^{-1}E_2^{-1}\cdots E_{N-1}^{-1}E_N^{-1} .$ ## Proof of $\Large{\det A=\det A^{\top}}$ If $A$ is not invertible, both determinants are zero.^[If $A^{\top}$ had inverse $B$, transposing $A^{\top}B=BA^{\top}=I$ gives $B^{\top}A=AB^{\top}=I$, making $A$ invertible—a contradiction.] Otherwise: write $A$ as a product of elementary matrices → transpose each factor and reverse their order → each factor keeps its determinant^[For swaps and scalings, $E^{\top}=E$. For row additions, $E$ and $E^{\top}$ are triangular with all diagonal entries $1$, so both determinants are $1$.] → multiply the same numbers in reverse order → $\det A=\det A^{\top}$. $ A=E_1E_2\cdots E_n \quad\Rightarrow\quad A^{\top}=E_n^{\top}\cdots E_2^{\top}E_1^{\top}. $ Thus we get $ \det A=\det A^{\top}. $ ## Proof of $\Large{\det(AB)=(\det A)(\det B)}$ If $B$ is invertible: write $B=E_1\cdots E_N$ → apply the elementary-factor determinant rule to $AB=AE_1\cdots E_N$ → multiply $\det A$ by all the elementary determinants → their product is $\det B$ ^[Invoke (1.5) onto $B$ such that we get $\det B=(\det E_{1})(\det E_{2})\ldots(\det E_{n})$. Then use (1.5) to get $AB$ into the format $(\det A)(\det E_{1})\ldots(\det E_{n})$ and we're done.]→ $\det(AB)=(\det A)(\det B)$. If $B$ is not invertible → $AB$ is not invertible → both sides are zero.